\newproblem{lay:7_2_27}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.2.27}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $A$ and $B$ be symmetric $n\times n$ matrices whose eigenvalues are all positive. Show that the eigenvalues of $A+B$ are all positive. [\textit{Hint}: Consider
	quadratic forms.]
}{
   % Solution
	If $A$ and $B$ are symmetric matrices, then $C=A+B$ is also symmetric. Consider now the quadratic form
	\begin{center}
		$Q_C(\mathbf{x})=\mathbf{x}^TC\mathbf{x}=\mathbf{x}^T(A+B)\mathbf{x}=\mathbf{x}^TA\mathbf{x}+\mathbf{x}^TB\mathbf{x}$
	\end{center}
	We may define the quadratic forms
	\begin{center}
		$Q_A(\mathbf{x})=\mathbf{x}^TA\mathbf{x}$\\
		$Q_B(\mathbf{x})=\mathbf{x}^TB\mathbf{x}$
	\end{center}
	So that $Q_C(\mathbf{x})=Q_A(\mathbf{x})+Q_B(\mathbf{x})$. Since $A$ and $B$ are symmetric matrices, these quadratic forms are well defined, and because all their eigenvalues
	are positive, then $Q_A$ and $Q_B$ are positive definite quadratic forms. This means that for any $\mathbf{x}\in\mathbb{R}^n$ it is verified that
	\begin{center}
		$Q_A(\mathbf{x})>0$\\
		$Q_B(\mathbf{x})>0$
	\end{center}
	Consequently, $Q_C(\mathbf{x})>0$, that is $Q_C$ is also positive definite and the eigenvalues of $C=A+B$ are all positive.
}
\useproblem{lay:7_2_27}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

